arithmetic sequences and series grade 12 caps...

Grade 12

CAPS

Mathematics

Series

Arithmetic

Sequences

and

Series

Outcomes for this Topic

In this topic you will:

• Investigate number sequences and generating

rules.

Unit 1

• Identify arithmetic sequences and construct the

formula for the general term.

Unit 2

• Discuss the arithmetic series and sigma notation.

Unit 3

• Construct and apply the sum formula of an arithmetic

series.

Unit 4

Grade 12

CAPS

Mathematics

Series

Unit 1

Number Sequences

and

Generating Rules

Number Sequences

DEFINITION: A number sequence is a list

of numbers given in a particular order.

Numbers in a sequence may be generated randomly

or by a particular rule

EXAMPLE 1: 1; 2; 4; 8; …

EXAMPLE 2: 18; 13; 8; 3; -2; …

EXAMPLE 3: 1; 4; 9; 16; …

EXAMPLE 4: 1; 8; 27; 64; …

Rule: Multiply by 2

Rule: Subtract 5 or add -5

Rule: Square a number

Rule: Cube of a number

Tutorial 1: Identifying Rules for Number

Sequences

• Identify the rule/pattern in the following sequences

• Write down the next three terms of each sequence

1. 5; 15; 45; … 2. 7; 4; 1; -2; …

3. 1; 1; 2; 3; 5; … 4. 2; 5; 10; 17; …

5. 27; 9; 3; 1; … 6. 0; 7; 26; 63; …

PAUSE

• Do Tutorial 1

• Then View Solutions

Tutorial 1 Problem 1: Suggested Solution

1. 5; 15; 45;

Generating Rule: Multiply by 3

Next three terms:

45 3 135

135 3 405

405 3 1 215


2. 7; 4; 1; 2;

Generating Rule: Add 3 or Subtract 3

Next three terms:

2 3 5

5 3 8

8 3 11


3. 1; 1; 2; 3; 5;

Generating Rule:

Add the two preceeding numbers

Next three terms:

3 5 8

8 5 13

13 8 21

1; 1; 2; 3; 5; 8;

1; 1; 2; 3; 5; 8; 13;

1; 1; 2; 3; 5; 8; 13; 21;

Fibonacci!!

Not required in

CAPS Syllabus


4. 2; 5; 10; 17;

Generating Rule:

Square Natural Numbers and Add 1

Next three terms:25 1 26

26 1 37 27 1 50


5. 27; 9; 3; 1;

Generating Rule:

1Divide by 3 or Multiply by

3

Next three terms:

11 3

3

1 1 1

3 3 9

1 1 1

9 3 27


6. 0; 7; 26; 63;

Generating Rule:

Cube the Natural Numbers minus 1

Next three terms:35 1 125 1 124 36 1 216 1 215 37 1 343 1 342

Grade 12 CAPS

Mathematics

Series

Unit 2

Arithmetic

Sequence

and the

General Term

Common Difference implies

an Arithmetic Sequence

The fixed number that we add is called the common

difference and is generally denoted by d.

Consider the following sequences:

2; 7; 12; 17; …

8; 2; -4; -10; …

6; 15; 24; 33; …

6 6 6

5 5 5

9 9 9 Common Difference

Arithmetic Sequence

n n 1Td T

2 1 3 2 4 3 T T T T T Td

Common Difference

General Term of an Arithmetic Sequence (AS)

Then the sequence is:

st

th

n

Let be the 1 term, the constant difference

and the n term of an Arithmetic SequT

da

ence

1 d0aT

2T 1a d

3T 2a d

10 a

T 9

d

nT (a n 1)d

thHence the general n term

of an AS is given by:

Explicit Form

a a a a; 2 3d; d; d; 4 d a ;

Problem 1 on Arithmetic Sequences

Hence the 19th term is -39 .

th

For a sequence 15; 12; 9; ... , determine

(a) the 11 term and

(b) which term is 39

Solution

Here 15 and a 3d

So15 ( 1)( 3) 39

15 3 3 39

3 57

19

n

n

n

n

b Let the term be 39.

Then ( 1) 39.

thn

na d

11a 10 15 (10)( 3) 15.daT

Problem 2 on Arithmetic Sequences

.

Solution

The 4th term of an AS is 14 and the 16th term is 50.

Calculate the first term and the common difference.

44 term : 3 14. (1) th dT a

1616 term : 15 50. (2) th dT a

Subtracting equation (1) from equation (2),

we get 12d d36 3

So 3(3) 14 5 a a

Recursive Formula for an Arithmetic

Sequence

Each new term in an arithmetic

sequence comes from adding

the common difference d to the

previous term. 1 a T

2 d a T

3 2d dT 2 Ta

10 9

T 9 T

da d

n

n 1

d

T (n 1)a

dT

Consider the terms

of an AS:

fHence a recursive ormula

for an AS is given by:

1

1 ; 1n n

T

n

a

T dT

Not required in

CAPS Syllabus

Tutorial 2: Questions on Arithmetic

Sequences

2; 4 ; 6 4

.

25; 14; 3

13th

x x x

x

1. are three consecutive terms of an AS.

Determine the value of

2. If ; ... forms an AS, determine

(a) the term,

(b) a formula for the general term,

(c) which term 52.is

1 1

4 20.

4

10

1 3 1

th

k kT T T k

3. The third term of an AS is and the seventh term is

Determine

(a) the first terms,

(b) the term

4. A sequence is defined by and for .

Determine the first four terms of the sequence.

PAUSE

• Do Tutorial 2

• Then View solutions


3 2 2 4 6x x x

Three consecutive terms are 8; 24 and 40

1. Given 2; 4 ; 6 4 as three consecutive

terms for an AS. Determine value of .

x x x

x

2 1 3 2For an AS, T T T T

(4 ) ( 2) (6 4) (4 )x x x x


2.(b) General term: ( 1)

25 ( 1)( 11)

36 11

nT a n d

n

n

Then 25 and 11.

2.(a) 13 term 12 25 12( 11) 107th

a d

a d

Hence the 8 term is 52.th

Given sequence 25;14; 3; as an AS:

2.(c) Suppose that 52.nT Then 36 11 52n

11 88 8.n n


(a) First four terms are: 4; 0; 4; 8.

(b) 10 term 9 4 9( 4) 32.th a d

3. 3 term 4 2 4 ... (1)

7 term 20 6 20 ... (2)

rd

th

a d

a d

Subtracting equation (1) from equation (2), we get

4 16 4

So 2( 4) 4 4.

d d

a a


1

2 1

3 2

4 3

4. 1

3 1 3 4

3 4 3 7

3 7 3 10

T

T T

T T

T T

1 11 3 1k kT T T k Given and for

Grade 12

CAPS

Mathematics

Series

Unit 3

Arithmetic Series

and

Sigma Notation

Arithmetic Series

1 2 n

n 1 2 n

T ;T ;...T

S T T ... T

Let denote the terms of an Arithmetic Sequence.

Then the sum of the terms of the sequence:

is Arithme called tic Se an ries

Definition

Example:

If we add a finite number of terms, the series is called finite.

nn SIf the first terms of an AS are added, the sum is denoted by .

4 5 9 13 17 44 S

7 5 9 13 17 21 25 29 119S

5; 9; 13; 17; 21;For the AS:

Sigma Notation and Series Expansion

1 2 n

n

1 2 n k

k 1

k

T T ... T

T T ... T T

T

A series can be written in as follows:

where is the general term of the c

Sigma N

orrespo

ota

ndi

t

n

io

g

n

sequence.

Example

5

1

(4 1) 4( ) 1 4( ) 1 4( ) 1 41 2 3 4 5( ) 1 4( ) 1

3 7 11 15 19 55

r

r

n

k k

k 1

T T , k 1 n.Read: the sum of the terms for to

62 2 2 2 2

3

3 4 5 6 9 16 25 36 86

k

k

Expanding a series from Sigma Notation

Choice of variables

, arbitrary r k

Writing a Series in Sigma Notation

To write a given series in sigma notation is more complicated.

For this k

we need the formula for the general term T .

Example

102 2 2 2 2 2

1

(a) Clearly for 1 to 10. Hence 1 2 3 ... 10

k

k

T k k k

5 5

1 1

Hence 5 11 17 23 29 (6 1)k

k k

T k

Write the series in Sigma Notation

2 2 2 2(a) 1 2 3 ... 10 (b) 5 11 17 23 29

Solution

(b) This is an AS with five terms and 5, 6.

1 5 1 6 6 1kT a k d k

a d

k

Tutorial 3: Arithmetic Series and Sigma Notation

4 5

1 0

5 201

2 1

( 1) (b) 2

(c) ( 1) (d) 5

k

k k

r

r k

k k

1. Expand and evaluate the following:

(a)

3 3 3 3

:

1 2 3 ... 7

( 3 4 5 ... 18

8 13 18 23 28

2. Write the following in sigma notation

(a)

b)

(c)

PAUSE

• Do Tutorial 3


Tutorial 3 Problem 1: Suggested Solutions

4

1

1. (a) ( 1) 1 2 2 3 3 4 4 5 40k

k k

20 20

0

1 1

(d) 5(1) 5( ) 5 5 5 ... 5 20 5 100k k

k

51

2

3 4 5 6

(c) ( 1)

1 1 1 1

1 1 1 1 0

r

r

50 1 2 3 4 5

0

(b) 2 2 2 2 2 2 2 1 2 4 8 16 32 63k

k


k

5

k 1

(c) This is an Arithmetic series with T 8 (k 1)5 5k 3;

where k 1 to 5. Hence 8 13 18 23 28 (5k 3).

3

k

73 3 3 3 3

k 1

2. (a) By inspection T k , for k 1 to 7.

Hence 1 2 3 ... 7 k .

k

18

k 3

16

k 1

(b) By inspection, T k, for k 3 to 18.

Hence 3 4 5 ... 18 k.

Alternative Solution : (k 2) (verify!!)

Grade 12

CAPS

Mathematics

Series

Unit 4

The Sum of

an Arithmetic

Series

Summing the terms of an Arithmetic Series

2 ... 2 1 ... (1) nS a a d a d a n d a n d

1 2 ... 2 ... (2) nS a n d a n d a d a d a

1 d.T aConsider an Arithmetic Series with and constant differen e c

Then

Reversing the order of terms in (1), we get:

2 2 1 2 1 ... 2 1 2 1nS a n d a n d a n d a n d

Summing the corresponding terms in (1) and (2) gives us:

2 2 ( 1) nS n a n d

Hence 2 12 n

nS a n d

Last term ( 1)

Hence 12

2

n

n

L n

nS a a n d

nS

a

a

d

L

Note

Note: Proof is examinable

Finding the sum of an Arithmetic Series

Example

5; 8; 11; 14; ...,

440.

For the Arithmetic sequence determine

(a) the sum of the first 30 terms,

(b) the number of terms that must be added for a sum total of

30

30(a) 2 5 29 3 15 97 1455

2 SSolution

2

(b) Suppose terms are added. Then 440.

2 5 1 3 4402

3 7 880 0

16 3 55 0

5516 and

3

nn S

nn

n n

n n

n n

16 Hence terms must be added.

Finding the Sum of an Arithmetic Series

Example

For the Arithmetic series 7 +11+15 + ...+151 determine

(a) the number of terms,

(b) the sum all the terms.

(a) Suppose the last term is the term. Then

7 1 4 151 7 4 4 151 4 148 37.

Hence there are 37 terms in the series.

th

n

n

T n n n n

Solution

37

37(b) The sum of all the terms is 7 151 2923.

2 S

Tutorial 4: Sums of Arithmetic Sequences

4. A job was advertised at a starting salary of

R90 000 pa with an annual increase of R4 500.

Determine:

(a) the employee's salary in the sixth year,

(b) the total earnings after 10 years.

1. Calculate the sum of:

(a) the first 18 terms of the sequence: 7; 13; 19; ...

(b) the first 7 terms of the sequence: 24; 21; 18; ...

2. Determine how many terms of the sequence 54; 51; 48; ...

must be added to make a sum of 510.

3. For a series 4 9 14 ... 99, determine:


(b) the sum of the terms.

PAUSE

• Do Tutorial 4



18

18 (a) 2 7 17 6 9(116) 1044

2S

1. Calculate the sum of:

(a) the first 18 terms of the sequence: 7; 13; 19; ...

(b) the first 7 terms of the sequence: 24; 21; 18; ...

7

7 7 (b) 2 24 6 3 30 105

2 2S


Hence summing either 17 or 20 terms will result in a sum of 510.

2. Determine how many terms of the sequence 54; 51; 48; ...

must be added to make a sum of 510.

If terms are added then 510.nn S

So 2 54 1 3 5102

nn

2

108 3 3 1020

3 111 1020 0

n n

n n

2 37 340 0

17 20 0

n n

n n

17 or 20n n

20

(b) Sum of the terms 4 99 1030.2


3. For a series 4 9 14 ... 99, determine:


(b) the sum of the terms.

(a) Suppose there are terms.

Then 4 1 5 99.n

n

T n

So 4 5 5 99 5 100 20.n n n Hence there are 20 terms in the series.

10

(b) Total earnings after 10 years:

10 2 90000 9 4500

2

1 102 500.

S

R


4. A job was advertised at a starting salary of

R90 000 pa with an annual increase of R4 500.

Determine:

(a) the employee's salary in the sixth year,

(b) the total earnings after 10 years.

6

(a) Salary in the sixth year:

90000 5(4500) 112 500.T R

End of the Topic Slides on Arithmetic

Sequences

REMEMBER!

• Consult text-books and past exam papers and memos

for additional examples.

• Attempt as many as possible other similar examples on

your own.

• Compare your methods with those that were discussed

in these Topic slides.

• Repeat this procedure until you are confident.

• Do not forget: Practice makes perfect!

arithmetic sequences and series grade 12 caps...

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